Question

sum of the digits of a two-digit number is 9 the number obtained by interchanging the digit exceeds the given by 27 find the original number​

Answer 1

Answer:

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Answer 2

Answer:

Let the two digit number be 10x+y.

Let the two digit number be 10x+y.Given that the sum of the digits is 9

Let the two digit number be 10x+y.Given that the sum of the digits is 9x+y=9

Let the two digit number be 10x+y.Given that the sum of the digits is 9x+y=9 Given that the number obtained by interchanging the digits exceeds the given number by 27

10y+x=10x+y+27

10y+x=10x+y+279x−9y=−27

10y+x=10x+y+279x−9y=−27On taking 9 as common

10y+x=10x+y+279x−9y=−27On taking 9 as commonx−y=−3

10y+x=10x+y+279x−9y=−27On taking 9 as commonx−y=−3 Adding equation (1) and (2)

10y+x=10x+y+279x−9y=−27On taking 9 as commonx−y=−3 Adding equation (1) and (2)x+y=9

10y+x=10x+y+279x−9y=−27On taking 9 as commonx−y=−3 Adding equation (1) and (2)x+y=9x−y=−3

10y+x=10x+y+279x−9y=−27On taking 9 as commonx−y=−3 Adding equation (1) and (2)x+y=9x−y=−32x=6

10y+x=10x+y+279x−9y=−27On taking 9 as commonx−y=−3 Adding equation (1) and (2)x+y=9x−y=−32x=6x=3

10y+x=10x+y+279x−9y=−27On taking 9 as commonx−y=−3 Adding equation (1) and (2)x+y=9x−y=−32x=6x=33+y=9

10y+x=10x+y+279x−9y=−27On taking 9 as commonx−y=−3 Adding equation (1) and (2)x+y=9x−y=−32x=6x=33+y=9y=6

10y+x=10x+y+279x−9y=−27On taking 9 as commonx−y=−3 Adding equation (1) and (2)x+y=9x−y=−32x=6x=33+y=9y=6Therefore, the number is 10x+y is 36.

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